$h(n) = -3n^{2}-2n-5(g(n))$ $f(t) = -5t^{2}-6t-h(t)$ $g(n) = 7n^{3}+n^{2}$ $ f(h(0)) = {?} $
First, let's solve for the value of the inner function, $h(0)$ . Then we'll know what to plug into the outer function. $h(0) = -3(0^{2})+(-2)(0)-5(g(0))$ To solve for the value of $h$ , we need to solve for the value of $g(0)$ $g(0) = 7(0^{3})+0^{2}$ $g(0) = 0$ That means $h(0) = -3(0^{2})+(-2)(0)+(-5)(0)$ $h(0) = 0$ Now we know that $h(0) = 0$ . Let's solve for $f(h(0))$ , which is $f(0)$ $f(0) = -5(0^{2})+(-6)(0)-h(0)$ To solve for the value of $f$ , we need to solve for the value of $h(0)$ $h(0) = -3(0^{2})+(-2)(0)-5(g(0))$ To solve for the value of $h$ , we need to solve for the value of $g(0)$ $g(0) = 7(0^{3})+0^{2}$ $g(0) = 0$ That means $h(0) = -3(0^{2})+(-2)(0)+(-5)(0)$ $h(0) = 0$ That means $f(0) = -5(0^{2})+(-6)(0)-0$ $f(0) = 0$